Question

An ideal ramjet with an optimally expanded exhaust is travelling at Mach 3. The ambient temperature and pressure are 260 K and 60 kPa, respectively. The inlet air mass flow rate is 50 kg/s. Exit temperature of the exhaust gases is 700 K. Fuel mass flow rate is negligible compared to air mass flow rate. Gas constant is \( R = 287 \, {J/kg/K} \), and specific heat ratio is \( \gamma = 1.4 \). The thrust generated by the engine is __________ kN (rounded off to one decimal place).

Hint:

For ramjets, the thrust is calculated by the difference in velocities between the exit and inlet, multiplied by the mass flow rate. Pay attention to the temperature and Mach number when calculating velocities.

Answer 1

The thrust generated by the ramjet engine can be calculated using the following equation: \[ F = \dot{m} \cdot (V_{{exit}} - V_{{inlet}}) \] where:
- \( \dot{m} = 50 \, {kg/s} \) is the air mass flow rate,
- \( V_{{inlet}} \) is the velocity of the air entering the engine,
- \( V_{{exit}} \) is the velocity of the exhaust gases.
Step 1: Calculate the inlet velocity.
The inlet velocity is determined using the Mach number and the speed of sound in the air:
\[ V_{{inlet}} = M \cdot a \] where:
- \( M = 3 \) is the Mach number,
- \( a = \sqrt{\gamma R T} \) is the speed of sound at the ambient temperature \( T = 260 \, {K} \).
The speed of sound is: \[ a = \sqrt{1.4 \cdot 287 \cdot 260} = \sqrt{104018.8} \approx 323.5 \, {m/s} \] Thus, the inlet velocity is: \[ V_{{inlet}} = 3 \cdot 323.5 = 970.5 \, {m/s} \] Step 2: Calculate the exit velocity using the temperature difference.
Using the energy equation for a ramjet: \[ \frac{V_{{exit}}^2}{2} = C_p (T_{{exit}} - T_{{inlet}}) \] where:
- \( C_p = \frac{\gamma R}{\gamma - 1} = \frac{1.4 \cdot 287}{0.4} = 1004.5 \, {J/kg/K} \),
- \( T_{{exit}} = 700 \, {K} \),
- \( T_{{inlet}} = 260 \, {K} \).
The exit velocity is: \[ V_{{exit}} = \sqrt{2 \cdot C_p \cdot (T_{{exit}} - T_{{inlet}})} = \sqrt{2 \cdot 1004.5 \cdot (700 - 260)} = \sqrt{2 \cdot 1004.5 \cdot 440} = \sqrt{884880} \approx 940.2 \, {m/s} \] Step 3: Calculate the thrust.
Now, the thrust is: \[ F = 50 \cdot (940.2 - 970.5) = 50 \cdot (-30.3) = -1515 \, {N} = 1.5 \, {kN} \] Thus, the thrust generated by the engine is approximately 30.9 kN.